Bài 2: Đúng ghi Đ, sai ghi S
a)\(\dfrac{48}{300}\)= 48%
b)\(\dfrac{48}{300}\) =\(\dfrac{16}{100}\) = 16%
c)\(\dfrac{30}{50}\)=\(\dfrac{60}{100}\)= 60%
d)\(\dfrac{30}{50}\)=\(\dfrac{60}{100}\)=6%
Bài 2: Đúng ghi Đ, sai ghi S
a)48/300= 48%
b)48/300 =16/100 = 16%
c)30/50=60/1000= 60%
d)30/50=60/100=6%
Đúng ghi Đ, sai ghi S
a) \(\dfrac{3}{10}< 0,3\) .......
\(\dfrac{3}{10}=0,3\) .......
b)\(\dfrac{135}{100}=1,35\) ....
\(\dfrac{135}{100}>1,35\) ........
c) 1\(\dfrac{7}{100}>1,7\) ......
1\(\dfrac{7}{100}< 1,7\)
Bài 1:
a) Phân số rút gọn được phân số tối giản \(\dfrac{2}{3}\) là:
A. \(\dfrac{12}{20}\) B. \(\dfrac{24}{48}\) C. \(\dfrac{34}{51}\) D. \(\dfrac{20}{60}\)
b) Tổng hai phân số là: \(\dfrac{9}{16}\). Nếu thêm vào số thứ nhất \(\dfrac{1}{4}\) thì tổng hai số là bao nhiêu?
A. \(\dfrac{1}{2}\) B. \(\dfrac{1}{3}\) C. \(\dfrac{1}{4}\) D. \(\dfrac{13}{16}\)
giúp mình với ạ, mình sẽ tick. Cảm ơn các bạn!
`->C`
`34/51= (34: 17)/(51:17)=2/3`
`->D`
`9/16 +1/4= 9/16+ 4/16=13/16`
a. C. \(\dfrac{34}{51}\)
b. D.\(\dfrac{13}{16}\)
Đúng ghi Đ sai ghi S
\(\dfrac{1}{4}\) = \(\dfrac{25}{100}\) ❏
\(\dfrac{3}{8}=\dfrac{35}{100}\) ❏
\(\dfrac{3}{8}=\dfrac{375}{1000}\) ❏
Chọn câu trả lời đúng
Rút gọn phân số \(\dfrac{48}{60}\) được phân số tối giản là:
A. \(\dfrac{24}{30}\) B. \(\dfrac{12}{15}\) C. \(\dfrac{3}{5}\) D. \(\dfrac{4}{5}\)
Đúng ghi Đ,sai ghi S:
a)\(\dfrac{1}{10}\) gấp 10 lần \(\dfrac{1}{100}\) __ b)\(\dfrac{1}{100}\) gấp 10 lần \(\dfrac{1}{10}\)__
c)\(\dfrac{1}{100}\) gấp lên 10 lần được \(\dfrac{1}{1000}\)__ d) \(\dfrac{1}{100}\) giảm đi 10 lần được \(\dfrac{1}{1000}\)__
__ là chỗ điền nha.
tính S=\(\dfrac{1}{30}+\dfrac{2}{48}+\dfrac{3}{88}+\dfrac{4}{165}+\dfrac{5}{300}\)
\(S=\dfrac{1}{30}+\dfrac{2}{48}+\dfrac{3}{88}+\dfrac{4}{165}+\dfrac{5}{300}\\ S=\dfrac{1}{5.6}+\dfrac{2}{6.8}+\dfrac{3}{8.11}+\dfrac{4}{11.15}+\dfrac{5}{15.20}\\ S=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}\\ S=\dfrac{1}{5}-\dfrac{1}{20}\\ S=\dfrac{4}{20}-\dfrac{1}{20}\\ S=\dfrac{3}{20}\)
Tính rồi viết kết quả dưới dạng số thập phân:
a) \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\) + \(\dfrac{16}{40}\) + \(\dfrac{25}{50}\) + \(\dfrac{36}{60}\) + \(\dfrac{49}{70}\) + \(\dfrac{64}{80}\) + \(\dfrac{81}{90}\)
b) ( \(\dfrac{4}{5}\) x \(\dfrac{3}{8}\) + \(\dfrac{4}{5}\) x \(\dfrac{5}{8}\) - \(\dfrac{4}{5}\) x \(\dfrac{7}{8}\) ) : \(\dfrac{1}{2}\)
\(a,=\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}=\dfrac{45}{10}=4,5\\ b,=\dfrac{4}{5}\times\left(\dfrac{3}{8}+\dfrac{5}{8}-\dfrac{7}{8}\right)\times2=\dfrac{8}{5}\times\dfrac{1}{8}=\dfrac{1}{5}=0,2\)
a) Rút gọn các phân số về tối giản, ta được:
\(\dfrac{1}{10}\)+\(\dfrac{2}{10}\)+\(\dfrac{3}{10}\)+\(\dfrac{4}{10}\)+\(\dfrac{5}{10}\)+\(\dfrac{6}{10}\)+\(\dfrac{7}{10}\)+\(\dfrac{8}{10}\)+\(\dfrac{9}{10}\)= kết quả là \(\dfrac{45}{10}\) ra số thập phân = \(4,5\)
b) \(\dfrac{4}{5}\) \(\times\) \(\left(\dfrac{3}{8}+\dfrac{5}{8}-\dfrac{7}{8}\right)\) = \(\dfrac{4}{5}\times\dfrac{1}{8}\) = \(\dfrac{4}{40}=\dfrac{1}{10}\)\(\div\dfrac{1}{2}\)
= \(\dfrac{2}{10}=0,2\)
CMR:
a) \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
b) \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}>48\)
a)\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{100^2-1}\)
\(A< \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}< \dfrac{50}{100}=\dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\)
b)B=\(\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{2499}{2500}\)
49-B=\(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
\(49-B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(49-B< 1-\dfrac{1}{50}< 1\Leftrightarrow49< 1+B\Leftrightarrow B>48\)(ĐPCM)
b) Đặt :
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+............+\dfrac{2499}{2500}\)
\(\Rightarrow A=\dfrac{4}{4}-\dfrac{1}{4}+\dfrac{9}{9}-\dfrac{1}{9}+.........+\dfrac{2500}{2500}-\dfrac{1}{2500}\)
\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...........+1-\dfrac{1}{50^2}\)
\(A=\left(1+1+....+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)\)(\(49\) chữ số \(1\))
\(A=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+........+\dfrac{1}{50^2}\right)\)
Lại có :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
Mà :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{50^2}\right)>49-1\)\(=48\)
\(\Rightarrow A>48\) \(\rightarrowđpcm\)